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shaqadry

  • 3 years ago

how to differentiate f(x) = ln ( 1 + √ x / 1 - √ x )

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  1. Jemurray3
    • 3 years ago
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    In general, \[ \frac{d}{dx} \ln( f(x) ) = \frac{1}{f(x)}\cdot f'(x)\]

  2. shaqadry
    • 3 years ago
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    how do i solve it? i cant seem to get the correct answer.

  3. Ahaanomegas
    • 3 years ago
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    Hint: You will have to use the chain rule by finding the derivative of the argument and multiplying it by the derivative that results without the chain rule. Are you aware of how to use the Chain Rule? If not, then I'll provide a solution.

  4. shaqadry
    • 3 years ago
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    im aware of it. but how do i calculate it?

  5. Ahaanomegas
    • 3 years ago
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    The derivative of the argument, you mean?

  6. shaqadry
    • 3 years ago
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    how do i apply chain rule in this equation

  7. ByteMe
    • 3 years ago
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    Rewrite f(x): \(\Large f(x)=ln\frac{1+\sqrt x}{1-\sqrt x}=ln(1+\sqrt x)-ln(1-\sqrt x) \) now just take the derivatives of each ln separately....

  8. Ahaanomegas
    • 3 years ago
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    Would you mind showing us your work? What have you found for the derivative of the argument of the natural log?

  9. shaqadry
    • 3 years ago
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    dy/dx = [ (1/1+√ x) (+ 1/2√x) ] - [ (1/1-√x) - ( 1/2√ x) ]

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